Let me just get Andrew Wiles for you

Question seen on Yahoo! Answers:
Is a^n+b^n=c^n when n>2 possible? plz give COMPLETE proof?
You'd think at some point in the process of looking at equations like these the teacher would have at least mentioned Fermat's Last Theorem.According to the Wiki article on the proof it is 150 pages long, so I declined to copy it into Y!A. Plus I'm not really sure the OP is up for elliptic curves and modular forms just yet.

Is fast internet possible? plz give COMPLETE proof?

150 pages long
I have a truly marvelous proof of this theorem, but this post box is too small to contain it.

I found a counterexample to the claim that all things must someday die, but I don't know how to show it to anyone.

150 pages long, so I declined to copy it into Y!A.
Just copy something from Mathgen. He'll never notice anyway.

Seems like a failed attempt to troll Yahoo answers. Sadly, both people who answered gave simple and straight answers. Oh...one of them was you.
At least you gave the integer qualifier.

You'd think at some point in the process of looking at equations like these the teacher would have at least mentioned Fermat's Last Theorem.
I read some scifi book a few years ago who suggested the question was silly, and rather than a^n + b^n = c^n, the equation should be written in the form
(Assuming I wrote that correctly. That is, for any n, you would write it a^n + b^n + c^n + d^n, so that there are n terms.) I got no idea if that's a more meaningful equality, just thought it was neat.

Is fast internet possible?
Not if you use MilwaukeePC internet.
plz give COMPLETE proof?
[insert list of all @ben_lubar's speedtest pictures here]

I got no idea if that's a more meaningful equality, just thought it was neat.
It's the same one, just rearranged and IMO obfuscated.

It's the same one, just rearranged and IMO obfuscated.
No, it's saying something a bit different. Original was a^n + b^n = c^n. This is more like asserting a^3 + b^3 + c^3 = d^3, or a^4 + b^4 + c^4 + d^4 = e^4, etc. You have one term per power.

Well, yes, obviously stuff varies if you change k (sorry, was thinking that but the editing process dropped it  not literally, @blakeyrat, just figuratively, meaning it didn't make it into the last post). It's the same when k = 2.
I suppose the book came up with a reason why that particular thing was interesting, because I'm not aware of any reason (which doesn't mean there isn't one).

I suppose the book came up with a reason why that particular thing was interesting, because I'm not aware of any reason (which doesn't mean there isn't one).
Not necessarily. Oh, I looked it up, it was Robert L Forward's Rocheworld series. An alien race simply suggests that generalizing a^2 + b^2 = c^2 to a^n + b^n = c^n without generalizing the number of terms was stupid. I thought the idea, if not necessarily the assertion, was interesting, without regard for whether it was meaningful, because it seems kind of elegant.

I thought the idea, if not necessarily the assertion, was interesting, without regard for whether it was meaningful, because it seems kind of elegant.
Indeed. It is the sort of thing that mathematicians do. But that particular equation is interesting because of the history. People were interested in finding out solutions to things like that, and obviously the Pythagorean theorem has been around for a long time, so it's a natural thing to speculate about. Plus, obviously, Fermat's Last Troll and who knows how many hours spent to prove it.


Assuming I wrote that correctly
The sigma should be of i = 1..n, not i = 1..k, if you want the number of terms to agree with the power.

An alien race simply suggests that generalizing a^2 + b^2 = c^2 to a^n + b^n = c^n without generalizing the number of terms was stupid.
If there is no solution for a^n + b^n = c^n, then there also isn't one for a^n + b^n + c^n = d^n etc (proof left as an exercise to the reader). That alien race was obviously not well versed in maths, or they have a very different idea of elegance. And peace.

Maybe not with 3 terms, but there is definitely a series of solution with \forall i a_i=1, b=2, k=2^n.

The sigma should be of i = 1..n, not i = 1..k, if you want the number of terms to agree with the power.
THANK YOU at least somebody read it carefully :
FWP: or ? So hard to choose....


Zoom.....
ENHANCE!
for less WTFy posting or to colourize and use as an avatar...

hmm... no i don't like that right edge at all....
that's better but the signature lost some definition.....
i need to find that PS7 cd it seems. PDN just isn't cutting it like it used to for this sort of CSI manipulation

@FrostCat said:
An alien race simply suggests that generalizing a^2 + b^2 = c^2 to a^n + b^n = c^n without generalizing the number of terms was stupid.
If there is no solution for a^n + b^n = c^n, then there also isn't one for a^n + b^n + c^n = d^n etc (proof left as an exercise to the reader). That alien race was obviously not well versed in maths, or they have a very different idea of elegance. And peace.
422481^4 == 95800^4 + 217519^4 + 414560^4
http://www.wolframalpha.com/input/?i=422481^4+%3D%3D+95800^4+%2B+217519^4+%2B+414560^4

Nice, but for the fourth power, you'd have to show
a^4 + b^4 + c^4 + d^4 = e^4.5 terms instead of 4 for teh 5th power, and so on.

That's the restriction where
n=k
. The generalization proposed had the number of terms independent of the power.

The generalization proposed had the number of terms independent of the power.
And someone pointed out that mistake already in it already.
EDIT: didn't have to scroll up as far as I thought.
@Maciejasjmj said:The sigma should be of i = 1..n, not i = 1..k, if you want the number of terms to agree with the power.

Yes, but apparently @FrostCat needed reminding.

Found these with a short python script:
1^1 == 1^1
3^2 + 4^2 == 5^2
3^3 + 4^3 + 5^3 == 6^3Then it started taking too long, so I found this: http://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture#Generalizations
30^4 + 120^4 + 272^4 + 315^4 == 353^4
19^5 + 43^5 + 46^5 + 47^5 + 67^5 == 72^5None known for n == 6.
127^7 + 258^7 + 266^7 + 413^7 + 430^7 + 439^7 + 525^7 == 568^7
90^8 + 223^8 + 478^8 + 524^8 + 748^8 + 1088^8 + 1190^8 + 1324^8 == 1409^8

Yes, but apparently @FrostCat needed reminding.
Nopethat's probably the most advanced math I'll use for years.

I find it interesting that such an academic question was posted on a Q&A site that gave the world How is babby formed?.

Oh don't worry, it's still about 95% stupid questions. I mostly stick to the Mathematics area to avoid the worst of it, but even then...
**Trending** 12 more that twice a number is equal to 48. What is the number? 25 answers
The sum of three consquative natural number is 39 what are that number?
25 answersSolve 2x+5=9?
1,283 answers... which is why I usually look for questions that nobody's answered yet, where there's at least a chance of getting something worthwhile.

"Consquative" is rather fine.

@some random yahoo said:
Is a^n+b^n=c^n when n>2 possible? plz give COMPLETE proof?
1^{3} + 0^{3} = 1^{3}

for less WTFy posting or to colourize and use as an avatar...
I don't know...I appreciate the effort, but the pasteitonatable, take a picture and upload seems like the kind of WTFy we'd want here in TRWTF land.
1^{3} + 0^{3} = 1^{3}
Violates a condition no one bothered to report here. From Wikipedia: "No three positive integers a, b, and c can satisfy the equation a^{n} + b^{n} = c^{n} for any integer value of n greater than two." Mathematicians do not consider 0 a positive integer.

Violates a condition no one bothered to report here.
If the idiot doesn't ask the question right, they deserve to get the trivial (and useless) answer.

They didn't specify natural numbers for any of the variables, so there are infinitely many answers

Violates a condition no one bothered to report here.
I did call it out in my actual answer on Y!A, because precision is important in mathematics.